So i dont seem to understand why i can't figure this out. it seemed like it would be simple enough. I am at a point in calc 2 where have the option of substitution pretty much. but i don.t seem where that option lies. I could work around to make it some inverse trig function but i haven't been able to figure that out either. So if you could help and show me the process i would be grateful. this has been bugging me for a while.
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∫5/(4 + 9r^2) dr
5/4 * ∫dr/(1 + (9r^2)/4)
Let z = 3r/2 => 2/3 dz = dr
5/6 * ∫dz/(1 + z^2) = 5/6 * arctan(z) + C
= 5/6 * arctan(3r/2) + C
5/4 * ∫dr/(1 + (9r^2)/4)
Let z = 3r/2 => 2/3 dz = dr
5/6 * ∫dz/(1 + z^2) = 5/6 * arctan(z) + C
= 5/6 * arctan(3r/2) + C
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r = (2/3)tan(t)